My Math Forum error: linear inequality (absolute function)

 Algebra Pre-Algebra and Basic Algebra Math Forum

 November 6th, 2017, 09:37 AM #1 Newbie   Joined: Nov 2017 From: Europe Posts: 2 Thanks: 0 error: linear inequality (absolute function) hello, I tried to solve the following inequality: abs(-2x+1)>2/3x+1 but get wrong results: -2x+1>-2/3x-1 I add 1 then +2x then multiply with 3 and divide by 4 I get: x<3/2 and -2x+1>2/3x+1 cancel out 1 then +2x 0>8/3x so I get x<0 surely this inequality is true for some x>0 - do you spot the error? best, richard Last edited by skipjack; November 6th, 2017 at 11:30 PM.
 November 6th, 2017, 06:01 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,859 Thanks: 1080 Math Focus: Elementary mathematics and beyond There are two cases; one where -2x + 1 is positive and one where it's negative. It's zero at x = 1/2. Does that help? Hint: make a sign chart - if you know how. If not, consult Google (or a search engine of your choice). You could also graph both functions on the same axes. Last edited by greg1313; November 6th, 2017 at 06:10 PM.
 November 7th, 2017, 04:07 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 Is the right side "(2/3)x+ 1" or "2/(3x)+ 1" or "2/(3x+ 1)"? Please use parentheses!
 November 8th, 2017, 07:56 AM #4 Newbie   Joined: Nov 2017 From: Europe Posts: 2 Thanks: 0 Hi Country Boy, sorry that's the one I meant: abs(-2x+1)>(2/3)x+1 Last edited by skipjack; November 8th, 2017 at 10:04 AM.
 November 8th, 2017, 10:20 AM #5 Global Moderator   Joined: Dec 2006 Posts: 19,542 Thanks: 1752 Let's first consider x < 1/2, so that abs(-2x + 1) = -2x + 1. To solve -2x + 1 > (2/3)x + 1, add 2x - 1 to both sides to get 0 > (8/3)x, which is equivalent to x < 0. So far, this matches what you did. Now let's consider x $\small\geqslant$ 1/2, so that abs(-2x + 1) = 2x - 1. To solve 2x - 1 > (2/3)x + 1, subtract (2/3)x - 1 from both sides to get (4/3)x > 2, which is equivalent to x > 3/2.
 November 10th, 2017, 03:09 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 By the way- this is NOT a "linear" inequality. Absolute value is not a linear function.

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